power ch552 with 5 or 3.3V

I am in the design process of a CH552 device which can be switched on 3.3V and 5V.ch552_pwr.png

I want to use a external 3.3V regulator to power V33 (pin16).?

If I understand it correctly Vcc rail powers the ports and flash prg circiut and V33 rail powers the core.

For normal operation this should? not be a problem since I can switch off the internal LDO.?

(bit bLDO3V3_OFF)


Question:

Will this work during bootloader mode? I am asking because during startup and bootmode both?

regulators will be powered.Therefore I added 10 Ohms in series to the external LDO.


Hello. I'm a Chinese user so text below may have some grammatical mistakes but just ignore them plz.

I think your schematic have no problem. Although WCH declared that CH552G can only be programmed while using 5V power supply, I have tested it can successfully be programmed using 3.3V power supply. I haven't tested it can be programmed definitely successfully for all situations so if you want to take it in mass production, use 5V power supply for programming.

Generally, if you want to use 5V to power CH552G, supply it to VCC(pin 15) and put 100nF both on VCC(pin 15) and V3.3(pin 16); If you want to use 3.3V to power CH552G, short VCC(pin 15) and V3.3(pin 16), then put 100nF on it.

The pull-up resistor to D+ is recommended to connect to VCC(pin 15), although while powered by 5V the voltage of D+ may exceed the limit but it's temporary pull-up to go into ISP mode so don't worry.

If you have any questions else, just reply this thread. For my personal advice, you can translate your text into Chinese and paste them in your reply together so WCH FAE can easily get your main idea.


Thanks for your reply. I already have done some designs with various CH55x chips but all with 5V.

My actual design requires switching between 5 and 3.3V. I am quite shure the above will work I just have some doubts about  connecting the regulator outputs together. In 5V mode V33 is a output and it is connected to my ext. LDO output. During powerup and bootmode I dont have control over the internal LDO so both LDO outputs drive each other.

BTW: connectiong the boot resistor to 3.3V is not a problem 3.3V is read as high even in 5V mode, so the bootloader can be activated. Done that many times.



Hello. I guess your concern is that if two LDOs are connected together, their small voltage difference may cause problems (such as high current)?

Connecting two LDOs directly in parallel can indeed cause problems. For example, if the voltage of LdoA is higher than that of LdoB, the feedback inside LdoB will cause it to almost completely turn off, which means that LdoA bears all the power at this time.

Returning to your situation, I personally think the worst-case scenario is that the internal LDO has a higher voltage than NCP1117, which will cause the internal LDO to bear all 3.3V current during the boot period. I am not sure how much 3.3V current is required in your design. If it does not exceed the output capacity of the internal LDO, I guess the above issue can be ignored.


Hello.


At the same time of 5V power supply, it is theoretically possible to input 3.3V on V33 pin. V33 pin is used as a reference reference for USB power supply and voltage, but you need to pay attention to that 3.3V line is best not connected to other external load devices. Excessive load may cause fluctuations of V33 pin and lead to USB or function abnormalities.


So V33 pin independence is usually recommended. An external LDO input can be used.



Hello. I got your concern.

The best solution is, if using 5V, disconnect external LDO from V33(pin 16), just let internal LDO power CH552G itself is okay but connecting external LDO through a resistor will also possibly work; if using 3.3V, connect external LDO to both VCC(pin 15) and V33(pin16).

Theoretically I think this should work fine for you as well.


Meanwhile i got my PCBs. Tests show that there is no problem with the switching beetween 3.3V and 5V operation. Even the 10R Resistor (R7) can savely replaced by a short or 0R.

There is no extra current or overload.


Thank you all for the support.


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